\newproblem{lay:4_1_1}{
  % Problem identification
	\begin{large}
	  \hspace{\fill}\newline
    \textbf{Lay, 4.1.1}
	\end{large}
	\\
  \ifthenelse{\boolean{identifyAuthor}}{\textit{Carlos Oscar Sorzano, Aug. 31st, 2013} \\}{}

  % Problem statement
	Let $V$ be the first quadrant in the $xy$-plane; that is, let
	\begin{center}
		$V=\left\{\begin{pmatrix}x\\y\end{pmatrix} \left| x,y\geq 0\right.\right\}$
	\end{center}
	\begin{enumerate}[a.]
		\item If $\mathbf{u}$ and $\mathbf{v}$ are in $V$, is $\mathbf{u}+\mathbf{v}$ in $V$? Why?
		\item Find a specific vector $\mathbf{u}\in V$ and a specific scalar $c$ such that $c\mathbf{u}$ is not in $V$. (This is enough to show that
					$V$ is not a vector space.)
	\end{enumerate}
}{
   % Solution
	\begin{enumerate}[a.]
		\item Let $\mathbf{u}=\begin{pmatrix}u_x\\u_y\end{pmatrix}$ and $\mathbf{v}=\begin{pmatrix}v_x\\v_y\end{pmatrix}$, then
		      \begin{center}
						$\mathbf{u}+\mathbf{v}=\begin{pmatrix}u_x\\u_y\end{pmatrix}+\begin{pmatrix}v_x\\v_y\end{pmatrix}=\begin{pmatrix}u_x+v_x\\u_y+v_y\end{pmatrix}$
					\end{center}
					If $u_x\geq 0$ and $v_x\geq 0$, then $u_x+v_x\geq 0$. Similarly for $u_y+v_y$. Consequently, $\mathbf{u}+\mathbf{v}$ is also in $V$.
		\item Let $V\ni \mathbf{u}=\begin{pmatrix}1\\0\end{pmatrix}$ and $c=-1$, then
		      \begin{center}
						$c\mathbf{u}=-\begin{pmatrix}1\\0\end{pmatrix}=\begin{pmatrix}-1\\0\end{pmatrix}$
					\end{center}
					that is not in $V$.
	\end{enumerate}
}
\useproblem{lay:4_1_1}
\ifthenelse{\boolean{eachProblemInOnePage}}{\newpage}{}
